If from the origin a chord is drawn to the circle `x^(2)+y^(2)-2x=0`, then the locus of the mid point of the chord has equation

Options

A. `x^(2)+y^(2)+x+y=0`
B. `x^(2)+y^(2)+2x+y=0`
C. `x^(2)+y^(2)-x=0`
D. `x^(2)+y^(2)-2x+y=0`

Answer

To find the locus of the midpoint of a chord drawn from the origin to the circle given by the equation \( x^2 + y^2 - 2x = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
The equation of the circle can be rewritten in standard form. We start with:
\[ x^2 + y^2 - 2x = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 = 0 \]
Completing the square for the \(x\) terms:
\[ (x - 1)^2 - 1 + y^2 = 0 \]
This simplifies to:
\[ (x - 1)^2 + y^2 = 1 \]
This represents a circle with center at \( (1, 0) \) and radius \( 1 \).

### Step 2: Midpoint of the Chord
Let \( (h, k) \) be the midpoint of the chord. The endpoints of the chord can be denoted as \( (p, q) \). According to the midpoint formula:
\[ h = \frac{0 + p}{2} \quad \text{and} \quad k = \frac{0 + q}{2} \]
This implies:
\[ p = 2h \quad \text{and} \quad q = 2k \]

### Step 3: Substitute into the Circle's Equation
Since the points \( (p, q) \) lie on the circle, we substitute \( p \) and \( q \) into the circle's equation:
\[ (2h - 1)^2 + (2k)^2 = 1 \]

### Step 4: Expand and Simplify
Expanding the equation:
\[ (2h - 1)^2 + (2k)^2 = 1 \]
This expands to:
\[ (4h^2 - 4h + 1) + 4k^2 = 1 \]
Combining like terms:
\[ 4h^2 + 4k^2 - 4h + 1 = 1 \]
Subtracting 1 from both sides gives:
\[ 4h^2 + 4k^2 - 4h = 0 \]

### Step 5: Factor and Rearrange
Factoring out 4:
\[ 4(h^2 + k^2 - h) = 0 \]
Dividing by 4:
\[ h^2 + k^2 - h = 0 \]
Rearranging gives:
\[ h^2 + k^2 = h \]

### Step 6: Replace \( h \) and \( k \) with \( x \) and \( y \)
Since \( h \) and \( k \) represent the coordinates of the midpoint, we can replace them with \( x \) and \( y \):
\[ x^2 + y^2 = x \]

### Final Equation
Thus, the locus of the midpoint of the chord is:
\[ x^2 + y^2 - x = 0 \]